Question
If $y = x^x$, find $\frac{\text{dy}}{\text{dx}}\text{at x}=\text{e}$
✓
Answer
We have, $y = x^x$ .....(i)
Taking log on both sides,
$\log\text{y}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
Putting x = e, we get,
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+\log_\text{e}\text{e})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+1)\big[\because\log_\text{e}\text{e}=1\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{e}^\text{e}$
Taking log on both sides,
$\log\text{y}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
Putting x = e, we get,
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+\log_\text{e}\text{e})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+1)\big[\because\log_\text{e}\text{e}=1\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{e}^\text{e}$
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